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b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
a)MTC 15
\(\dfrac{\left(x-3\right)\times3}{15}=\dfrac{6.15-\left(1-2x\right)\times5}{15}=\dfrac{3x-9}{15}=\dfrac{90-5-10x}{15}=3x-9=90-5-10x\Leftrightarrow3x+10x=90-5+9\)
Chưa nghỉ tết à :))
\(a,\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)
\(\Rightarrow3\left(x-3\right)=6.15-5\left(1-2x\right)\)
\(\Leftrightarrow3x-9=90-5+10x\)
\(\Leftrightarrow3x-10x=90-5+9\)
\(\Leftrightarrow-7x=94\)
\(\Leftrightarrow x=-\dfrac{94}{7}\)
Vậy.....
\(b,\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Rightarrow2\left(3x-2\right)-5.12=3\left[3-2\left(x+7\right)\right]\)
\(\Leftrightarrow6x-4-60=-6x-33\)
\(\Leftrightarrow6x+6x=-33+60+4\)
\(\Leftrightarrow12x=31\)
\(\Leftrightarrow x=\dfrac{31}{12}\)
Vậy.....
\(c,2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)\)
\(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
\(\Leftrightarrow2x+x=5-\dfrac{13}{5}-\dfrac{6}{5}\)
\(\Leftrightarrow3x=\dfrac{6}{5}\)
\(\Leftrightarrow x=\dfrac{2}{5}\)
Vậy.....
\(d,\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
\(\Rightarrow28\left[5\left(x-1\right)+2\right]-42\left(7x-1\right)=24\left[2\left(2x+1\right)\right]-5.168\)
\(\Leftrightarrow140x-84-294x+42=96x+48-840\)
\(\Leftrightarrow140x-294x-96x=48-840-42+84\)
\(\Leftrightarrow-250x=-750\)
\(\Leftrightarrow x=3\)
Vậy.....
\(e,\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)
\(\Rightarrow6\left(x-1\right)+3\left(x-1\right)=12-4\left[2\left(x-1\right)\right]\)
\(\Leftrightarrow6x-6+3x-3=12-8x+8\)
\(\Leftrightarrow6x+3x+8x=12+8+3+6\)
\(\Leftrightarrow17x=29\)
\(\Leftrightarrow x=\dfrac{29}{17}\)
Vậy.....
\(g,\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
\(\Leftrightarrow\dfrac{2}{2001}-\dfrac{x}{2001}-1=\dfrac{1}{2002}-\dfrac{x}{2002}-\dfrac{x}{2003}\)
\(\Leftrightarrow-\dfrac{x}{2001}+\dfrac{x}{2002}+\dfrac{x}{2003}=\dfrac{1}{2002}+1-\dfrac{2}{2001}\)
\(\Leftrightarrow x\left(-\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}\right)=1+\dfrac{1}{2002}-\dfrac{2}{2001}\)
\(\Leftrightarrow x=\dfrac{\left(1+\dfrac{1}{2002}-\dfrac{2}{2001}\right)}{\left(-\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}\right)}=2003\)
Vậy.....
a ) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)(1)
ĐKXĐ : \(x\ne1;x\ne2\)
(1)\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow2-x+5x+5=15\)
\(\Leftrightarrow4x+7=15\\\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\left(KTMĐKXĐ\right)\)
Vậy pt vô nghiệm .
b ) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( 2 )
ĐKXĐ : \(x\ne3;x\ne-2\)
(2) \(\Leftrightarrow3x-x^2+6-2x+x^2+2x=3x+6-x^2-2x\)
\(\Leftrightarrow x^2+2x=0\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐKXĐ\right)\\x=-2\left(KTMĐKXĐ\right)\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S={0}.
c ) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) (3)
ĐKXĐ : \(x\ne1;x\ne3\)
\(\left(3\right)\Leftrightarrow\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)
\(\Leftrightarrow6\left(3-x\right)+4\left(x-1\right)=8\)
\(\Leftrightarrow18-6x+4x-4=8\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\)
Vậy tập nghiệm của phương trình là S={-3}
d ) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (4)
ĐKXĐ : \(x\ne0;x\ne2\)
\(\left(4\right)\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTMĐKXĐ\right)\\x=-1\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S={-1}
a) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\) ( đk: x ≠ -1; x ≠ 2 )
\(\Leftrightarrow\) \(\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow\) \(2-x+5\left(x+1\right)=15\)
\(\Leftrightarrow\) \(2-x+5x+5=15\)
\(\Leftrightarrow\)\(4x=8\)
\(\Rightarrow\) \(x=2\) ( KTM )
S = ∅
b) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( đk: x ≠ - 2 ; x ≠ 3 )
\(\Leftrightarrow\) \(\left(x+2\right)\left(3-x\right)+x\left(x+2\right)=5x+2\left(3-x\right)\)
\(\Leftrightarrow\) \(3x-x^2+6-2x+x^2+2x=5x+6-2x\)
\(\Leftrightarrow\) \(3x+6=3x+6\)
\(\Rightarrow\)\(0x=0\) ( TM )
\(\Rightarrow\) Phương trình vô số nghiệm
S = R
c) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) ( đk: x ≠ 1 ; x ≠ 3 )
\(\Leftrightarrow\) \(\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)
\(\Leftrightarrow\)\(6\left(3-x\right)+4\left(x-1\right)=8\)
\(\Leftrightarrow\) \(18-6x+4x-4=8\)
\(\Leftrightarrow\) \(-2x=-6\)
\(\Rightarrow x=3\) ( KTM )
S = ∅
d) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (đk: x ≠ 2; x ≠ 0 )
\(\Leftrightarrow\) \(x\left(x+2\right)-x+2=2\)
\(\Leftrightarrow\) \(x^2+2x-x+2=2\)
\(\Leftrightarrow\) \(x^2+x=0\)
\(\Leftrightarrow\) \(x\left(x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\left(KTM\right)\\x=1\left(TM\right)\end{matrix}\right.\)
S = \(\left\{2\right\}\)